The commutant and invariant subspaces for dual truncated Toeplitz operators

Abstract

Dual truncated Toeplitz operators on the orthogonal complement of the model space $K_u^2(=H^2 \ominus uH^2)$ with u nonconstant inner function are defined to be the compression of multiplication operators to the orthogonal complement of $K_u^2$ in $L^2$. In this paper, we give a complete characterization of the commutant of dual truncated Toeplitz operator $D_z$, and we even obtain the commutant of all dual truncated Toeplitz operators with bounded analytic symbols. Moreover, we describe the nontrival invariant subspaces of $D_z$.

1 Introduction

Let $H^2$ be the classical Hardy space of open unit disk ${\mathbb {D}}=\{z \in {\mathbb {C}}: |z|<1\}$ and $L^2=L^2({\mathbb {T}})$ denote the usual Lebesgue space on the unit circle ${\mathbb {T}}=\{z \in {\mathbb {C}}: |z|=1\}$. Let $H^{\infty }$ be the space of bounded analytic functions on ${\mathbb {D}}$. By Fatou’s theorem and harmonic extension [12], we usually identify $H^2$ with the closed subspace of $L^2({\mathbb {T}})$ consisting of the functions with vanishing fourier coefficients with negative indices. Let P denote the orthogonal projection from $L^2$ to $H^2.$ For $\varphi \in L^{\infty }$, the Toeplitz operator $T_{\varphi }: H^2 \rightarrow H^2$ is defined by

$$\begin{aligned} T_{\varphi }x=P(\varphi x), x \in H^2. \end{aligned}$$

Let $P_{-}=I-P$ denote the orthogonal projection from $L^2$ to $(H^2)^{\bot }=\overline{zH^{2}}$, the Hankel operator $H_{\varphi }: H^2 \rightarrow (H^2)^{\bot }$ is defined by

$$\begin{aligned} H_{\varphi }x=P_{-}(\varphi x), x \in H^2 \end{aligned}$$

and an easy calculation gives

$$\begin{aligned} H_{\varphi }^{*}y=P({\bar{\varphi }}y), y \in (H^2)^{\bot }. \end{aligned}$$

The dual Toeplitz operator $S_{\varphi }$ defined on $(H^2)^{\bot }$ is given by

$$\begin{aligned} S_{\varphi }x=P_{-}(\varphi x), x \in (H^2)^{\bot }. \end{aligned}$$

For $f \in L^2$, define $Vf(z)={\bar{z}}\overline{f(z)}$; then V is an anti-unitary operator on $L^2$ and satisfies

$$\begin{aligned} V=V^{-1}, \, VT_f=S_{{\bar{f}}}V. \end{aligned}$$

Let u be a nonconstant inner function. $K_u^2=H^2 \ominus uH^2$ is called the model space, which is an invariant subspace for $T_{{\bar{z}}}$ [2]. Let $P_u$ denote the orthogonal projection from $L^2$ to $K_u^2$. For $\varphi \in L^2$, the truncated Toeplitz operator $A_{\varphi }$, which was first introduced by Sarason [19], is defined on $K_u^2$ by

$$\begin{aligned} A_{\varphi }x=P_u(\varphi x), x \in K_u^2. \end{aligned}$$

The truncated Hankel operator $B_{\varphi }: K_u^2 \rightarrow (K_u^2)^{\bot }$ is defined by

$$\begin{aligned} B_{\varphi }x=(I-P_u)(\varphi x), x \in K_u^2, \end{aligned}$$

and it is easy to check that

$$\begin{aligned} B_{\varphi }^{*}y=P_u({\bar{\varphi }} x), \, y \in (K_u^2)^{\bot }. \end{aligned}$$

The dual truncated Toeplitz operator $D_{\varphi }$ on the orthogonal complement of $K_u^2$, which was introduced first by X. Ding and Y. Sang in [9] and independently introduced by Y. Hu etc. in [13], is defined by

$$\begin{aligned} D_{\varphi }y=(I-P_u)(\varphi y), y \in (K_u^2)^{\bot }. \end{aligned}$$

It is known that a Toeplitz operator on $H^2$ is anti-unitarily equivalent to multiplication on $(H^2 )^{\bot }$ followed by projection onto $(H^2 )^{\bot }$, but it is no longer the case in the Bergman space setting. Stroethoff and Zheng [24] studied the algebraic and spectral properties of dual Toeplitz operators on the orthogonal complement of Bergman space in $L^2({\mathbb {D}}, dA)$ and detected to what extent these properties parallel those of Toeplitz operators on the Hardy space. Recently, Ding-Sang [9] and Sang-Qin-Ding [22, 23] obtained a series of interesting results about dual truncated Toeplitz operators, which are dual Toeplitz operators of truncated Toeplitz operators on $K_{u}^2$. Some of their results are similar to the classical operator theory of Toeplitz operators on the Hardy space, but others are pretty different. Motivated by the above work, in this paper, we mainly study the commutant and invariant subspaces of dual truncated Toeplitz operator $D_{z}$. Our results for dual truncated Toeplitz operators may offer some insight into the study of similar questions for dual Toeplitz operators on the model space.

Let $M_{\varphi }$ denote the multiplication operator on $L^2$; then on one hand, on the decomposition $L^2=H^2 \oplus (H^2)^{\bot }$, a standard calculation shows that

$$\begin{aligned} M_{\varphi }=\left[ \begin{array}{ccccc} T_{\varphi } &{} H_{{\bar{\varphi }}}^{*} \\ H_{\varphi } &{} S_{\varphi } \end{array}\right] , \end{aligned}$$

on the other hand, if we consider the decomposition $L^2=K_u^2 \oplus (K_u^2)^{\bot }$, then

$$\begin{aligned} M_{\varphi }=\left[ \begin{array}{ccccc} A_{\varphi } &{} B_{{\bar{\varphi }}}^{*} \\ B_{\varphi } &{} D_{\varphi } \end{array}\right] . \end{aligned}$$

Recall that the commutant of an operator T on a Hibert space ${\mathcal {H}}$ is the collection $\{T\}^{'}$ consisting of operators which commute with T. According to the operator theory of classical Hardy Toeplitz operators [3, Theorem 7], for unilateral shift $T_z$ on Hardy space,

$$\begin{aligned} \{T_z\}^{'}=\{T_{\varphi }: \varphi \in H^{\infty }({\mathbb {T}})\}, \end{aligned}$$

where $H^{\infty }({\mathbb {T}})$ is the collection of bounded functions on ${\mathbb {T}}$ whose negative Fourier coefficients vanish. The commutant of the Bergman shift $\mathrm {T}_z$ [6] on Bergman space $L_a^{2}$ has the following description:

$$\begin{aligned} \{\mathrm {T}_z\}^{'}=\{\mathrm {T}_{\varphi }: \varphi \in H^{\infty }\}. \end{aligned}$$

There is an identification $H^{\infty }({\mathbb {T}}) \cong H^{\infty }$ via the Poisson integral and Fatou’s theorem.

One would expect an analogous result for dual truncated Toeplitz operator $D_z$ on the orthogonal complement of $K_u^2$; however, unlike the commutants of Hardy shift $T_z$ and Bergman shift $\mathrm {T}_z$, not all the operators in $\{D_z\}^{'}$ are dual truncated Toeplitz operators. Using the techniques in the operator theory of Toeplitz operators on classical Hardy space, we obtain the following result about the commutant of dual truncated Toeplitz operator $D_z$:

Theorem 1

Let u be a nonconstant inner function.

(1) If $u(0)=0$, then

$$\begin{aligned} \{D_{z}\}^{'}=\left\{ \left[ \begin{array}{ccccc} t_{\varphi } &{} 0 \\ b_{{\bar{u}}f} &{} S_{\psi } \end{array}\right] : \varphi , \psi \in H^{\infty }, \, f \in L^{\infty } \right\} , \end{aligned}$$

where the operator matrix is written with respect to the decomposition $(K_u^2)^{\bot }=uH^2 \oplus \overline{zH^{2}},$ $t_{\varphi }x=M_uPM_{{\bar{u}}}(\varphi x) \text {for} \, x \in uH^2$ is called a little Toeplitz operator, and $b_{f{\bar{u}}}x=P_{-}(f{\bar{u}}x) \text {for} \, x \in uH^2$ is called a little Hankel operator.

(2) If $u(0) \ne 0$, then

$$\begin{aligned} \{D_{z}\}^{'}=\left\{ \left[ \begin{array}{ccccc} t_{\varphi } &{} {\bar{u}}(0)b_{\overline{u\varphi }}^{*}\\ \frac{1}{{\bar{u}}(0)}b_{{\bar{u}}\varphi } &{} S_{\varphi } \end{array}\right] : \varphi \in L^{\infty } \right\} . \end{aligned}$$

On the one hand, for Toeplitz operator $T_g$ with $g \in H^{\infty }$ on the Hardy space [10],

$$\begin{aligned} \{T_g: g \in H^{\infty }\}^{'}=\{ T_{\varphi }: \varphi \in H^{\infty }\}, \end{aligned}$$

and $\{ T_{\varphi }: \varphi \in H^{\infty }\}$ is a commutative Banach algebra, while an easily constructed example in [23] showed that $\{D_g: g \in H^{\infty }\}$ is not commutative nor an algebra. On the other hand, according to [23, Corollary 1], for $\varphi \in K_u^2 \cap H^{\infty }$,

$$\begin{aligned} \{D_{\psi }: \psi \in K_u^2 \cap H^{\infty }\} \subseteq \{D_{\varphi }\}^{'}, \end{aligned}$$

which indicates that $\{D_{\varphi }\}^{'}$ with $\varphi \in H^{\infty }$ is quite large. Hence it is interesting to know which operators are included in $\{D_{\varphi }: \varphi \in H^{\infty }\}^{'}$.

Taking the above theorem a step further, we describe the commutant of the set consisting of all dual truncated Toeplitz operators with bounded analytic symbols, which is quite different from the Toeplitz operator theory on Hardy space.

Theorem 2

Let u be a nonconstant inner function. Then

$$\begin{aligned} \{D_g: g \in H^{\infty }\}^{'}=\left\{ cI: \, c \in {\mathbb {C}} \right\} , \end{aligned}$$

where I is the indentity operator on $(K_u^2)^{\bot }$.

One of the outstanding unsolved problems in operator theory is the invariant subspace problem: Does every bounded linear operator on a separable infinite dimensional Hilbert space have a nontrivial invariant closed subspace? Read [18] and P. Enflo [11] settled the problem for Banach spaces, they obtained Banach spaces and operators without having nontrivial invariant subspaces. Instead of making counterexamples, some people started characterizing the invariant subspaces of certain classes of operators. Beurling’s theorem [2] is the first significant step in this direction, which says that ${\mathcal {M}} \subset H^2$ is a nontrivial closed invariant subspace of $T_z$ if and only if ${\mathcal {M}}=\varphi H^2$ for some inner function $\varphi$. Sarason [20] characterized all closed invariant subspaces of the real Volterra operator on $L^2[0, 1]$ by using Beurling’s result. And he also studied the invariant subspaces of multiplication plus real Volterra operator on $L^p[0, 1]$ in [21]. Subsequently, Ong [15] studied invariant subspaces of the shift plus a multiple of the Volterra operator on $L^2[0, 1].$ Aleman and Korenblum [1] characterized the invariant subspaces of the complex Volterra acting on the Hardy space. More generally, C̆uc̆ković and Paudyal described the invariant subspaces of the shift plus a complex Volterra operator in [7] [8]. For an overview on the invariant subspace problem, we refer the monograph by Radjavi and Rosenthal [17] or the more recent book by Chalendar and Partington [4].

Motivated by Beurling’s work on the invariant subspaces of Toeplitz operator $T_z$ on Hardy space, in this paper, we give a characterization of the invariant subspaces of dual truncated Toeplitz operator $D_z$ by using classical techniques in the Toeplitz operator theory on the Hardy space.

Theorem 3

Let u be a nonconstant inner function with $u(0)=0$ and ${\mathcal {M}}$ be a nontrivial closed subspace of $(K_u^2)^{\bot }$. Then one of the following cases happens:

• (a) If ${\mathcal {M}} \subset uH^2$, then ${\mathcal {M}}$ is an invariant subspace of $D_z$ if and only if there exists a nonconstant inner function $\theta$ such that

  $$\begin{aligned} {\mathcal {M}}=\theta uH^2; \end{aligned}$$

• (b) If ${\mathcal {M}} \subset (H^2)^{\bot }$, then ${\mathcal {M}}$ is an invariant subspace of $D_z$ if and only if there exists a nonconstant inner function $\eta$ such that

  $$\begin{aligned} {\mathcal {M}}={\bar{z}} \overline{K_{\eta }^2}; \end{aligned}$$

• (c) Let ${\mathcal {M}} \nsubseteq uH^2$ and ${\mathcal {M}} \nsubseteq (H^2)^{\bot }$. If ${\mathcal {M}}$ is an invariant subspace of $D_z$, then there exist nonconstant inner functions $\theta$ and $\eta$ such that $cl(uP{\bar{u}}{\mathcal {M}})=\theta uH^2$ and $cl(P_{-}{\mathcal {M}})={\bar{z}} \overline{K_{\eta }^2},$ where $cl({\mathcal {N}})$ denotes the closure of ${\mathcal {N}}$. Conversely, if there exist inner functions $\theta$ and $\eta$ with one of them nonconstant such that ${\mathcal {M}}_{+}=cl(uP{\bar{u}}{\mathcal {M}})=\theta uH^2$ and ${\mathcal {M}}_{-}=cl(P_{-}{\mathcal {M}})={\bar{z}} \overline{K_{\eta }^2}$, then ${\mathcal {M}}_{+} \oplus {\mathcal {M}}_{-}$ is a nontrivial invariant subspace of $D_z$.

If u is a nonconstant inner function with $u(0) \ne 0$, then the structure of invariant subspaces for $D_z$ is described as follows:

Theorem 4

Let u be a nonconstant inner function with $u(0) \ne 0$. Then the nontrivial closed subspace ${\mathcal {M}}$ of $(K_u^2)^{\bot }$ is an invariant subspace of $D_z$ if and only if one of the following cases happens:

• (a) there exists a unimodular function $\varphi$ such that

  $$\begin{aligned} {\mathcal {M}}=U_A( \varphi H^2), \end{aligned}$$

  where $U_A=\left[ \begin{array}{ccccc} M_u&{}0\\ 0&{}\frac{1}{{\bar{u}}(0)}I_{\overline{zH^2}} \end{array}\right]$ on $L^2=H^2 \oplus \overline{zH^2}$;

• (b) there exists a Borel set $E \subset {\mathbb {T}}$ with $0< m_E <1$ such that

  $$\begin{aligned} {\mathcal {M}}=U_AL_{E}^2, \end{aligned}$$

  where $m_E$ is the Lebesgue measure of E and $L_E^2=\{f \in L^2: f|_{{\mathbb {T}}-E}=0\}$.

2 Preliminaries

In this section, we mainly review some elementary facts and recent results of dual truncated Toeplitz operators.

Suppose u is a nonconstant inner function; then an easy calculation shows that the orthogonal projection $Q_u$ from $L^2$ to $(K_u^2)^{\bot }$ can be written as $Q_u=M_uPM_{{\bar{u}}}+P_{-}$. Define an operator $U=\left[ \begin{array}{ccccc} M_u&{} 0\\ 0 &{} I_{\overline{zH_2}} \end{array}\right]$ on $L^2=H^2 \oplus (H^2)^{\bot }$; then U is a unitary operator from $L^2$ to $(K_u^2)^{\bot }=uH^2 \oplus (H^2)^{\bot }$, and clearly, $U^{*}=\left[ \begin{array}{ccccc} M_{{\bar{u}}}&{} 0\\ 0 &{} I_{\overline{zH_2}} \end{array}\right]$.

It is not hard to obtain the following useful matrix representation of $D_{\varphi }$, which shows that the dual truncated Toeplitz operators are closely related to the Toeplitz operators and Hankel opertators on $H^2$ and truncated Toeplitz operators.

Lemma 1

([23]) If $\varphi \in L^{\infty }$, then on $L^2=H^2 \oplus \overline{zH^2}$,

$$\begin{aligned} U^{*}D_{\varphi }U=\left[ \begin{array}{ccccc} T_{\varphi }&{} H_{u{\bar{\varphi }}}^{*}\\ H_ {u\varphi }&{} S_{\varphi } \end{array}\right] . \end{aligned}$$

Remark 1

In the following, we denote ${\widetilde{D}}_{\varphi }:=U^{*}D_{\varphi }U$ for the sake of simplicity. Especially, from the above lemma, if $\varphi \in H^{2}$, then

$$\begin{aligned} {\widetilde{D}}_{\varphi }=U^{*}D_{\varphi }U=\left[ \begin{array}{ccccc} T_{\varphi }&{} H_{u{\bar{\varphi }}}^{*}\\ 0&{} S_{\varphi } \end{array}\right] . \end{aligned}$$

For $\varphi \in L^{\infty }$, $D_{\varphi }$ has a $2 \times 2$ matrix representation on $(K_u^2)^{\bot }=uH^2 \oplus \overline{zH^2}$ as

$$\begin{aligned} D_{\varphi }=\left[ \begin{array}{ccccc} t_{\varphi }&{} b_{{\bar{\varphi }}}^{*}\\ b_{\varphi }&{} S_{\varphi } \end{array}\right] , \end{aligned}$$

where $t_{\varphi }: uH^2 \rightarrow uH^2$ is defined by

$$\begin{aligned} t_{\varphi }x=M_uPM_{{\bar{u}}}(\varphi x), \, \forall x \in uH^2, \end{aligned}$$

and we call $t_{\varphi }$ a little Toeplitz operator in the following: $b_{\varphi }: uH^2 \rightarrow \overline{zH^2}$ is defined by

$$\begin{aligned} b_{\varphi }x=P_{-}\varphi x, \, \forall x \in uH^2. \end{aligned}$$

Obviously $b_{\varphi }=H_{\varphi }|_{uH^2}$, so it is reasonable that we call $b_{\varphi }$ a little Hankel operator.

Since $M_u$ is a unitary operator from $H^2$ to $uH^2$, for any $x \in H^2$, $ux \in uH^2$, we have

$$\begin{aligned} t_{\varphi }ux=uP{\bar{u}}\varphi ux=uP\varphi x; \end{aligned}$$

then $M_{{\bar{u}}}t_{\varphi }M_ux=P\varphi x=T_{\varphi }x;$ hence $M_{{\bar{u}}}t_{\varphi }M_u=T_{\varphi },$ that is, $t_{\varphi }$ is unitarily equivalent to $T_{\varphi }$.

Furthermore, it is easy to obtain an operator $b_{\varphi }^{*}: \overline{zH^2} \rightarrow uH^2$ given by

$$\begin{aligned} b_{\varphi }^{*}y=uP{\bar{u}}{\bar{\varphi }}y=M_uH_{u\varphi }^{*}y, \, \forall y \in \overline{zH^2}, \end{aligned}$$

that is, $b_{\varphi }^{*}=M_uH_{u\varphi }^{*}.$

With the above notations in our mind, especially, for $\varphi \in H^{\infty },$ since $b_{\varphi }=0,$ we have

$$\begin{aligned} D_{\varphi }=\left[ \begin{array}{ccccc} t_{\varphi }&{} b_{{\bar{\varphi }}}^{*}\\ 0&{} S_{\varphi } \end{array}\right] . \end{aligned}$$

Recently, Y. Qin, the second and the third authors ([9, 22, 23])have obtained a series of interesting results about dual truncated Toeplitz operators. Some of their results on dual truncated Toeplitz operators are similar to the classical operator theory of Toeplitz operators on the Hardy space, but others are pretty different. Here we list a few of them as follows:

Theorem 5

([9]) Let $\varphi \in L^2;$ then

(1) $D_{\varphi }$ is bounded on $(K_u^2)^{\bot }$ if and only if $\varphi \in L^{\infty }$ and $\Vert D_{\varphi }\Vert =\Vert \varphi \Vert _{\infty }$.

(2) $D_{\varphi }$ is compact if and only if $\varphi =0.$

Theorem 6

([9, 23]) Let $\varphi , \psi \in L^{\infty },$ and u be an inner function.

• (1) If $D_{\varphi }$ is invertible, then $\varphi$ is invertible in $L^{\infty }$.

• (2) $D_{\varphi }D_{\psi }=0$ if and only if $\varphi$ or $\psi$ is zero.

• (3) $D_{\varphi }D_{\psi }=D_{\varphi \psi }$ if and only if one of the following cases holds:

• (a) $\varphi , \psi , {\bar{\varphi }}(u-\lambda ), {\bar{\psi }}(u-\lambda )$ and $\overline{\varphi \psi }(u-\lambda )$ all belong to $H^2$ for some constant $\lambda .$

• (b) ${\bar{\varphi }}, {\bar{\psi }}, \varphi (u-\lambda ), \psi (u-\lambda )$ and $\varphi \psi (u-\lambda )$ all belong to $H^2$ for some constant $\lambda .$

• (c) Either $\varphi$ or $\psi$ is constant.

• (4) $D_{\varphi }D_{\psi }=D_{\psi }D_{\varphi }$ if and only if one of the following cases holds:

• (a) $\varphi , \psi , {\bar{\varphi }}(u-\lambda )$ and ${\bar{\psi }}(u-\lambda )$ all belong to $H^2$ for some constant $\lambda .$

• (b) ${\bar{\varphi }}, {\bar{\psi }}, \varphi (u-\lambda )$ and $\psi (u-\lambda )$ all belong to $H^2$ for some constant $\lambda .$

• (c) A nontrival linear combination of $\varphi$ and $\psi$ is constant.

Suppose $\mathfrak {D}_{\mathfrak {X}}=clos\{\sum \nolimits _{i=1}^{n}\prod \nolimits _{j=1}^{m}D_{\varphi _{ij}}:\varphi _{ij} \in \mathfrak {X}\}$, where $\mathfrak {X}$ is a closed self-adjoint subalgebra of $L^{\infty }$; then $\mathfrak {D}_{\mathfrak {X}}$ is a $C^{*}$-subalgebra of $B((K_u^2)^{\bot })$. Y. Qin, the second and the third authors obtained the following two short exact sequences:

Theorem 7

([22]) (1) The sequence

$$\begin{aligned} 0 \rightarrow \mathfrak {GD}_{L^{\infty }} \rightarrow \mathfrak {D}_{L^{\infty }} \rightarrow L^{\infty } \rightarrow 0 \end{aligned}$$

is a short exact sequence, where $\mathfrak {GD}_{L^{\infty }}$ is the semicommutator ideal of $\mathfrak {D}_{L^{\infty }}.$

(2) The sequence

$$\begin{aligned} 0 \rightarrow {\mathcal {K}} \rightarrow \mathfrak {D}_{C({\mathbb {T}})} \rightarrow C({\mathbb {T}}) \rightarrow 0 \end{aligned}$$

is a short exact sequence, where ${\mathcal {K}}$ denotes the set of all compact operators on $(K_u^2)^{\bot }.$

3 The commutant of dual truncated Toeplitz operators

In this section, we shall study the commutant of $D_z$ first, and further we characterize the commutant of the set of all dual truncated Toeplitz operators with bounded analytic symbols. We obtain some results which are quite different from the operator theory of Toeplitz operators both on classical Hardy spaces and Bergman spaces.

Definition 1

Let $B({\mathcal {H}})$ be the set of all bounded linear operators on the Hilbert space ${\mathcal {H}}$ and $S \subset B({\mathcal {H}})$, the commutant of S is denoted by $S^{'}$ and is defined to be the set of all operators in $B({\mathcal {H}})$ that commute with each operator in S.

The following is the first main theorem of this section:

Theorem 1

Let u be a nonconstant inner function.

(1) If $u(0)=0$, then

$$\begin{aligned} \{D_{z}\}^{'}=\left\{ \left[ \begin{array}{ccccc} t_{\varphi } &{} 0 \\ b_{{\bar{u}}f} &{} S_{\psi } \end{array}\right] : \varphi , \psi \in H^{\infty }, \, f \in L^{\infty } \right\} , \end{aligned}$$

where the operator matrix is written with respect to the decomposition $(K_u^2)^{\bot }=uH^2 \oplus \overline{zH^{2}}$.

(2) If $u(0) \ne 0$, then

$$\begin{aligned} \{D_{z}\}^{'}=\left\{ \left[ \begin{array}{ccccc} t_{\varphi } &{} {\bar{u}}(0)b_{\overline{u\varphi }}^{*}\\ \frac{1}{{\bar{u}}(0)}b_{{\bar{u}}\varphi } &{} S_{\varphi } \end{array}\right] : \varphi \in L^{\infty } \right\} . \end{aligned}$$

Proof

Suppose $X \in \{D_z\}^{'}$, ${\widetilde{X}}=U^{*}XU$ is a bounded linear operator on $L^2$, where

$$\begin{aligned} U=\left[ \begin{array}{ccccc} M_u&{} 0\\ 0 &{} I_{\overline{zH_2}} \end{array}\right] \end{aligned}$$

is a unitary operator from $L^2=H^2 \oplus (H^2)^{\bot }$ to $(K_u^2)^{\bot }=uH^2 \oplus (H^2)^{\bot }$. Then

$$\begin{aligned} {\widetilde{D}}_z=U^{*}D_zU=\left[ \begin{array}{ccccc} T_z&{} H_{u{\bar{z}}}^{*}\\ 0 &{}S_z \end{array}\right] \end{aligned}$$

Since $XD_z=D_zX$, we have

$$\begin{aligned} {\widetilde{D}}_z{\widetilde{X}}={\widetilde{X}}{\widetilde{D}}_z. \end{aligned}$$

Assume

$$\begin{aligned} {\widetilde{X}}=\left[ \begin{array}{ccccc} X_{11}&{}X_{12}\\ X_{21} &{}X_{22} \end{array}\right] , \end{aligned}$$

where the operator matrix is written with respect to the decomposition $L^2=H^2 \oplus (H^2)^{\bot }$, that is, $X_{11}: H^2 \rightarrow H^2$, $X_{12}: (H^2)^{\bot } \rightarrow H^2,$ $X_{21}: H^2 \rightarrow (H^2)^{\bot }$ and $X_{22}: (H^2)^{\bot } \rightarrow (H^2)^{\bot }$; then

$$\begin{aligned} \left[ \begin{array}{ccccc} T_{z}&{}H_{u{\bar{z}}}^{*}\\ 0&{}S_{z} \end{array}\right] \left[ \begin{array}{ccccc} X_{11}&{}X_{12}\\ X_{21} &{}X_{22} \end{array}\right] =\left[ \begin{array}{ccccc} X_{11}&{}X_{12}\\ X_{21} &{}X_{22} \end{array}\right] \left[ \begin{array}{ccccc} T_{z}&{}H_{u{\bar{z}}}^{*}\\ 0&{}S_{z} \end{array}\right] , \end{aligned}$$

that is,

$$\begin{aligned} \left[ \begin{array}{ccccc} T_{z}X_{11}+H_{u{\bar{z}}}^{*}X_{21}&{}T_{z}X_{12}+H_{u{\bar{z}}}^{*}X_{22}\\ S_{z}X_{21}&{}S_{z}X_{22} \end{array}\right] =\left[ \begin{array}{ccccc} X_{11}T_{z}&{}X_{11}H_{u{\bar{z}}}^{*}+X_{12}S_z\\ X_{21}T_z &{}X_{21}H_{u{\bar{z}}}^{*}+X_{22}S_z \end{array}\right] . \end{aligned}$$

Thus we get the following equations:

$$\begin{aligned} T_{z}X_{11}+H_{u{\bar{z}}}^{*}X_{21}&=X_{11}T_{z} \end{aligned}$$

(1)

$$\begin{aligned} T_{z}X_{12}+H_{u{\bar{z}}}^{*}X_{22}&=X_{11}H_{u{\bar{z}}}^{*}+X_{12}S_z\end{aligned}$$

(2)

$$\begin{aligned} S_{z}X_{21}&=X_{21}T_z \end{aligned}$$

(3)

$$\begin{aligned} S_{z}X_{22}&=X_{21}H_{u{\bar{z}}}^{*}+X_{22}S_z \end{aligned}$$

(4)

(1) If $u(0)=0$, then $u{\bar{z}} \in H^2$; thus we have $H_{u{\bar{z}}}=0$, and then $H_{u{\bar{z}}}^{*}=0$. That is, ${\widetilde{D}}_z$ is a diagonal operator in this case. Therefore, from the above equations, we get

$$\begin{aligned} T_{z}X_{11}&=X_{11}T_{z} \end{aligned}$$

(5)

$$\begin{aligned} T_{z}X_{12}&=X_{12}S_z \end{aligned}$$

(6)

$$\begin{aligned} S_{z}X_{21}&=X_{21}T_z \end{aligned}$$

(7)

$$\begin{aligned} S_{z}X_{22}&=X_{22}S_z \end{aligned}$$

(8)

First, using the fact [3, Theorem 7] that

$$\begin{aligned} \{T_z\}^{'}=\{ T_{\varphi }: \varphi \in H^{\infty }\}, \end{aligned}$$

Equation (5) implies that there exists some $\varphi \in H^{\infty }$ such that $X_{11}=T_{\varphi }$.

Second, as $X_{12}$ is a bounded linear operator from $\overline{zH^{2}}$ to $H^2$, by Eq. (6), we have

$$\begin{aligned} T_zX_{12}{\bar{z}}=X_{12}S_z{\bar{z}}=X_{12}P_{-}z{\bar{z}}=X_{12}P_{-}1=0. \end{aligned}$$

Since $\ker T_z=\{0\}$, we obtain $X_{12}{\bar{z}}=0$. Furthermore, for $n \ge 2$, using Eq. (6) again, then

$$\begin{aligned} T_zX_{12}{\bar{z}}^n=X_{12}S_z{\bar{z}}^n=X_{12}{\bar{z}}^{n-1}. \end{aligned}$$

By induction, for all $n \ge 1$, we get,

$$\begin{aligned} X_{12}{\bar{z}}^n=0, \end{aligned}$$

thus $X_{12}=0.$

Third, Eq. (7) indicates that $X_{21}$ is a bounded Hankel operator by Theorems 1.1 and 1.8 in [16]; thus there is some $f \in L^{\infty }$ such that $X_{21}=H_{f}$.

Finally by Eq. (8) and the properties of the anti-unitary operator V, we have

$$\begin{aligned} VS_{z}VVX_{22}V=VX_{22}VVS_zV; \end{aligned}$$

then,

$$\begin{aligned} T_{{\bar{z}}}(VX_{22}V)=(VX_{22}V)T_{{\bar{z}}}, \end{aligned}$$

we get

$$\begin{aligned} T_{z}(VX_{22}V)^{*}=(VX_{22}V)^{*}T_{z}. \end{aligned}$$

According to [3, Theorem 7],

$$\begin{aligned} \{T_z\}^{'}=\{ T_{\varphi }: \varphi \in H^{\infty }\}, \end{aligned}$$

$VX_{22}V$ is a bounded Toeplitz operator, so there exists some $\psi \in H^{\infty }$, such that $VX_{22}V=T_{\overline{\psi }}$; thus

$$\begin{aligned} X_{22}=VT_{\overline{\psi }}V=S_{\psi }, \psi \in H^{\infty }. \end{aligned}$$

It follows that

$$\begin{aligned} {\widetilde{X}}=\left[ \begin{array}{ccccc} T_{\varphi }&{}0\\ H_{f} &{}S_{\psi } \end{array}\right] , \end{aligned}$$

where $\varphi , \psi \in H^{\infty }$, and $f \in L^{\infty }$. Hence, on $(K_u^2)^{\bot }=uH^2 \oplus \overline{zH^{2}}$, we have

$$\begin{aligned} X&=U{\widetilde{X}}U^*\\&=\left[ \begin{array}{ccccc} M_{u}&{}0\\ 0&{}I_{\overline{zH_2}} \end{array}\right] \left[ \begin{array}{ccccc} T_{\varphi }&{}0\\ H_{f}&{}S_{\psi } \end{array}\right] \left[ \begin{array}{ccccc} M_{{\bar{u}}}&{}0\\ 0&{}I_{\overline{zH_2}} \end{array}\right] \\&=\left[ \begin{array}{ccccc} M_uT_{\varphi }M_{{\bar{u}}}&{}0\\ H_{f}M_{{\bar{u}}} &{}S_{\psi } \end{array}\right] \\&=\left[ \begin{array}{ccccc} t_{\varphi }&{}0\\ b_{f{\bar{u}}}&{}S_{\psi } \end{array}\right] . \end{aligned}$$

Therefore, if $X \in \{D_z\}^{'}$, then X has the matrix representation as above.

Conversely, if

$$\begin{aligned} X=\left[ \begin{array}{ccccc} t_{\varphi }&{}0\\ b_{f{\bar{u}}}&{}S_{\psi } \end{array}\right] , \end{aligned}$$

where $\varphi , \psi \in H^{\infty }$, and $f \in L^{\infty }$, we need to show that $XD_z=D_zX.$ Since

$$\begin{aligned} D_z=\left[ \begin{array}{ccccc} t_{z}&{}b^{*}_{{\bar{z}}}\\ 0&{}S_{z} \end{array}\right] , \end{aligned}$$

note that $u(0)=0$, for any $y \in \overline{zH^{2}}$, we have

$$\begin{aligned} b^{*}_{{\bar{z}}}y=uP{\bar{u}}zy=uP({\bar{u}}z)y=0, \end{aligned}$$

it follows that

$$\begin{aligned} D_z=\left[ \begin{array}{ccccc} t_{z}&{}0\\ 0&{}S_{z} \end{array}\right] . \end{aligned}$$

Then

$$\begin{aligned} XD_z=\left[ \begin{array}{ccccc} t_{\varphi }&{}0\\ b_{f{\bar{u}}}&{}S_{\psi } \end{array}\right] \left[ \begin{array}{ccccc} t_{z}&{}0\\ 0&{}S_{z} \end{array}\right] =\left[ \begin{array}{ccccc} t_{\varphi }t_{z}&{}0\\ b_{f{\bar{u}}}t_{z}&{}S_{\psi }S_{z} \end{array}\right] =\left[ \begin{array}{ccccc} t_{z\varphi }&{}0\\ b_{f{\bar{u}}}t_{z}&{}S_{z\psi } \end{array}\right] . \end{aligned}$$

On the other hand,

$$\begin{aligned} D_zX=\left[ \begin{array}{ccccc} t_{z}&{}0\\ 0&{}S_{z} \end{array}\right] \left[ \begin{array}{ccccc} t_{\varphi }&{}0\\ b_{f{\bar{u}}}&{}S_{\psi } \end{array}\right] =\left[ \begin{array}{ccccc} t_{z\varphi }&{}0\\ S_zb_{f{\bar{u}}}&{}S_{z\psi } \end{array}\right] . \end{aligned}$$

By the straightforward calculation, for any $x \in H^2, ux \in uH^2$, we have

$$\begin{aligned} b_{f{\bar{u}}}t_{z}ux&=b_{f{\bar{u}}}uP{\bar{u}}zux=b_{f{\bar{u}}}uPzx\\&=P_{-}f{\bar{u}}uPzx=P_{-}fPzx=H_{f}T_zx, \end{aligned}$$

and

$$\begin{aligned} S_zb_{f{\bar{u}}}ux=S_zP_{-}f{\bar{u}}ux=S_zP_{-}fx=S_zH_{f}x. \end{aligned}$$

Since $S_zH_{f}=H_{f}T_z$, it follows that

$$\begin{aligned} b_{f{\bar{u}}}t_{z}=S_zb_{f{\bar{u}}}. \end{aligned}$$

Therefore, we obtain $XD_z=D_zX.$

Combining the above two directions, we get

$$\begin{aligned} \{D_{z}\}^{'}=\left\{ \left[ \begin{array}{ccccc} t_{\varphi } &{} 0 \\ b_{{\bar{u}}f} &{} S_{\psi } \end{array}\right] : \varphi , \psi \in H^{\infty }, \, f \in L^{\infty } \right\} . \end{aligned}$$

(2) If $u(0) \ne 0$, then

$$\begin{aligned} D_z=\left[ \begin{array}{ccccc} t_{z}&{}b_{{\bar{z}}}^{*}\\ 0&{}S_{z} \end{array}\right] , \end{aligned}$$

and

$$\begin{aligned} {\widetilde{D}}_z=U^{*}D_zU=\left[ \begin{array}{ccccc} T_{z}&{}H_{{\bar{z}}u}^{*}\\ 0&{}S_{z} \end{array}\right] . \end{aligned}$$

Note that $(u-u(0)){\bar{z}} \in H^2$, hence $H_{(u-u(0)){\bar{z}}}=0$, it follows that

$$\begin{aligned} H_{u{\bar{z}}}=u(0)H_{{\bar{z}}}; \end{aligned}$$

thus

$$\begin{aligned} H_{u{\bar{z}}}^{*}={\bar{u}}(0)H_{{\bar{z}}}^{*}. \end{aligned}$$

So we have

$$\begin{aligned} {\widetilde{D}}_z=U^{*}D_zU=\left[ \begin{array}{ccccc} T_{z}&{}{\bar{u}}(0)H_{{\bar{z}}}^{*}\\ 0&{}S_{z} \end{array}\right] . \end{aligned}$$

Let $A=\left[ \begin{array}{ccccc} I_{H_2}&{}0\\ 0&{}\frac{1}{{\bar{u}}(0)}I_{\overline{zH_2}} \end{array}\right]$ on $L^2=H^2 \oplus \overline{zH^{2}}$; then $A^{-1}=\left[ \begin{array}{ccccc} I_{H_2}&{}0\\ 0&{}{\bar{u}}(0)I_{\overline{zH_2}} \end{array}\right] .$ Since $M_z=\left[ \begin{array}{ccccc} T_z&{}H_{{\bar{z}}}^{*}\\ 0&{}S_z \end{array}\right] ,$ a straightforward calculation gives that

$$\begin{aligned} AM_z=\left[ \begin{array}{ccccc} I_{H_2}&{}0\\ 0&{}\frac{1}{{\bar{u}}(0)}I_{\overline{zH_2}} \end{array}\right] \left[ \begin{array}{ccccc} T_z&{}H_{{\bar{z}}}^{*}\\ 0&{}S_z \end{array}\right] =\left[ \begin{array}{ccccc} T_z&{}H_{{\bar{z}}}^{*}\\ 0&{}\frac{1}{{\bar{u}}(0)}S_z \end{array}\right] , \end{aligned}$$

and

$$\begin{aligned} {\widetilde{D}}_zA=\left[ \begin{array}{ccccc} T_{z}&{}{\bar{u}}(0)H_{{\bar{z}}}^{*}\\ 0&{}S_{z} \end{array}\right] \left[ \begin{array}{ccccc} I_{H_2}&{}0\\ 0&{}\frac{1}{{\bar{u}}(0)}I_{\overline{zH_2}} \end{array}\right] =\left[ \begin{array}{ccccc} T_z&{}H_{{\bar{z}}}^{*}\\ 0&{}\frac{1}{{\bar{u}}(0)}S_z \end{array}\right] . \end{aligned}$$

Hence $AM_z={\widetilde{D}}_zA$, it follows that ${\widetilde{D}}_z=AM_zA^{-1}.$ On the other hand, suppose $B \in \{{\widetilde{D}}_z\}^{'}$, that is,

$$\begin{aligned} B{\widetilde{D}}_z={\widetilde{D}}_zB, \end{aligned}$$

which holds if and only if

$$\begin{aligned} BAM_zA^{-1}=AM_zA^{-1}B, \end{aligned}$$

equivalently,

$$\begin{aligned} (A^{-1}BA)M_z=M_z(A^{-1}BA); \end{aligned}$$

hence equivalently,

$$\begin{aligned} A^{-1}BA \in \{M_z\}^{'}, \end{aligned}$$

which means,

$$\begin{aligned} B \in A\{M_z\}^{'}A^{-1}. \end{aligned}$$

Since $\{M_z\}^{'}=\{M_{\varphi }: \varphi \in L^{\infty }\}$(see [10, 4.22 Proposition]), it follows that

$$\begin{aligned} \{{\widetilde{D}}_z\}^{'}=A\{M_z\}^{'}A^{-1}=\{AM_{\varphi }A^{-1}: \varphi \in L^{\infty }\}. \end{aligned}$$

By a simple calculation, we get

$$\begin{aligned} AM_{\varphi }A^{-1}&=\left[ \begin{array}{ccccc} I_{H_2}&{}0\\ 0&{}\frac{1}{{\bar{u}}(0)}I_{\overline{zH_2}} \end{array}\right] \left[ \begin{array}{ccccc} T_{\varphi }&{}H_{{\bar{\varphi }}}^{*}\\ H_{\varphi }&{}S_{\varphi } \end{array}\right] \left[ \begin{array}{ccccc} I_{H_2}&{}0\\ 0&{}{\bar{u}}(0)I_{\overline{zH_2}} \end{array}\right] \\&=\left[ \begin{array}{ccccc} T_{\varphi }&{}{\bar{u}}(0)H_{{\bar{\varphi }}}^{*}\\ \frac{1}{{\bar{u}}(0)}H_{\varphi }&{}S_{\varphi } \end{array}\right] . \end{aligned}$$

Therefore,

$$\begin{aligned} \{{\widetilde{D}}_z\}^{'}=\left\{ \left[ \begin{array}{ccccc} T_{\varphi }&{}{\bar{u}}(0)H_{{\bar{\varphi }}}^{*}\\ \frac{1}{{\bar{u}}(0)}H_{\varphi }&{}S_{\varphi } \end{array}\right] : \varphi \in L^{\infty } \right\} . \end{aligned}$$

Note that ${\widetilde{D}}_z=U^{*}D_zU$, then $D_z=U{\widetilde{D}}_zU^{*}$; thus we have

$$\begin{aligned} \{D_z\}^{'}=\{U{\widetilde{D}}_zU^{*}\}^{'}=\left\{ U\left[ \begin{array}{ccccc} T_{\varphi }&{}{\bar{u}}(0)H_{{\bar{\varphi }}}^{*}\\ \frac{1}{{\bar{u}}(0)}H_{\varphi }&{}S_{\varphi } \end{array}\right] U^{*}: \varphi \in L^{\infty } \right\} . \end{aligned}$$

Furthermore,

$$\begin{aligned} U\left[ \begin{array}{ccccc} T_{\varphi }&{}{\bar{u}}(0)H_{{\bar{\varphi }}}^{*}\\ \frac{1}{{\bar{u}}(0)}H_{\varphi }&{}S_{\varphi } \end{array}\right] U^{*}= & {} \left[ \begin{array}{ccccc} M_u&{}0\\ 0&{}I_{\overline{zH_2}} \end{array}\right] \left[ \begin{array}{ccccc} T_{\varphi }&{}{\bar{u}}(0)H_{{\bar{\varphi }}}^{*}\\ \frac{1}{{\bar{u}}(0)}H_{\varphi }&{}S_{\varphi } \end{array}\right] \left[ \begin{array}{ccccc} M_{{\bar{u}}}&{}0\\ 0&{}I_{\overline{zH_2}} \end{array}\right] \\= & {} \left[ \begin{array}{ccccc} M_uT_{\varphi }M_{{\bar{u}}}&{}{\bar{u}}(0)M_uH_{{\bar{\varphi }}}^{*}\\ \frac{1}{{\bar{u}}(0)}H_{\varphi }M_{{\bar{u}}}&{}S_{\varphi } \end{array}\right] \\= & {} \left[ \begin{array}{ccccc} t_{\varphi }&{}{\bar{u}}(0)b_{\overline{u\varphi }}^{*}\\ \frac{1}{{\bar{u}}(0)}b_{\varphi {\bar{u}}}&{}S_{\varphi } \end{array}\right] . \end{aligned}$$

Therefore,

$$\begin{aligned} \{D_z\}^{'}=\left\{ \left[ \begin{array}{ccccc} t_{\varphi }&{}{\bar{u}}(0)b_{\overline{u\varphi }}^{*}\\ \frac{1}{{\bar{u}}(0)}b_{\varphi {\bar{u}}}&{}S_{\varphi } \end{array}\right] : \varphi \in L^{\infty } \right\} . \end{aligned}$$

This completes the proof. $\square$

In the following, we write $T \in \{D_1, D_2, \dots , D_n\}^{'}$ if $TD_1=D_1T, TD_2=D_2T, \dots ,$ and $TD_n=D_nT$, where $T, D_1, D_2, \dots , D_n$ are bounded linear operators on the same Hilbert space. Taking the above result a step further, we obtain the commutant of $\{D_g: g \in H^{\infty }\}$, which is pretty different from the case of Toeplitz opertator theory on Hardy space. A direct calculation shows that, in general, $D_{z}D_{z} \ne D_{z^2}$ on $(K_{u}^{2})^{\bot }$ with $u\ne z^{2}u_{0},$ where $u_{0}$ is inner . So we study the commutant of $\{D_{z^n}: n \in {\mathbb {N}}\}$ first.

Lemma 2

Let u be a nonconstant inner function. Then

$$\begin{aligned} \{D_{z^n}: n \in {\mathbb {N}}\}^{'}=\left\{ cI: \, c \in {\mathbb {C}} \right\} , \end{aligned}$$

where I is the identity operator on $(K_u^2)^{\bot }$.

Proof

Note that cI with c constant always commutes with $D_{z^n}$ with $n \in {\mathbb {N}}$; thus we only need to prove that

$$\begin{aligned} \{D_{z^n}: n \in {\mathbb {N}}\}^{'} \subset \left\{ cI: \, c \in {\mathbb {C}} \right\} . \end{aligned}$$

(9)

Assume $T \in \{D_{z^n}: n \in {\mathbb {N}}\}^{'}$; then we have $TD_{z^{n}}=D_{z^{n}}T$ for any positive integer n. Suppose u is a nonconstant inner function with Fourier series $u(z)=\sum \nolimits _{m=0}^{+ \infty } a_mz^m$. We use $a_k$ to denote the first nonzero coefficient of u. In the following, we will prove (9) in two cases: $k = 0$ and $k \geqslant 1$.

(a) For the case $k = 0$, that is, $a_0 \ne 0$, since $TD_{z}=D_{z}T$, by Theorem 1, there exists some $\varphi \in L^{\infty }$ with Fourier series $\varphi (z)=\sum \nolimits _{m=1}^{+ \infty } c_{-m}{\bar{z}}^m+\sum \nolimits _{m=0}^{+ \infty } c_mz^m$ such that

$$\begin{aligned} T=\left[ \begin{array}{ccccc} t_{\varphi } &{} {\bar{a}}_0b_{\overline{u\varphi }}^{*}\\ \frac{1}{{\bar{a}}_0}b_{{\bar{u}}\varphi } &{} S_{\varphi } \end{array}\right] . \end{aligned}$$

Since

$$\begin{aligned} D_{z^{n}}=\left[ \begin{array}{ccccc} t_{z^{n}}&{} b_{{\bar{z}}^{n}}^{*}\\ 0&{} S_{z^{n}} \end{array}\right] , \end{aligned}$$

and $TD_{z^{n}}=D_{z^{n}}T \; (n \geqslant 1)$, it follows that

$$\begin{aligned} \left[ \begin{array}{ccccc} t_{\varphi } &{} {\bar{a}}_0b_{\overline{u\varphi }}^{*}\\ \frac{1}{{\bar{a}}_0}b_{{\bar{u}}\varphi } &{} S_{\varphi } \end{array}\right] \left[ \begin{array}{ccccc} t_{z^{n}}&{} b_{{\bar{z}}^{n}}^{*}\\ 0&{} S_{z^{n}} \end{array}\right] =\left[ \begin{array}{ccccc} t_{z^{n}}&{} b_{{\bar{z}}^{n}}^{*}\\ 0&{} S_{z^{n}} \end{array}\right] \left[ \begin{array}{ccccc} t_{\varphi } &{} {\bar{a}}_0b_{\overline{u\varphi }}^{*}\\ \frac{1}{{\bar{a}}_0}b_{{\bar{u}}\varphi } &{} S_{\varphi } \end{array}\right] . \end{aligned}$$

Therefore, we get the following equations:

$$\begin{aligned} t_{\varphi }t_{z^{n}}&=t_{z^{n}}t_{\varphi }+\frac{1}{{\bar{a}}_0}b_{{\bar{z}}^{n}}^{*}b_{{\bar{u}}\varphi }, \end{aligned}$$

(10)

$$\begin{aligned} t_{\varphi }b_{{\bar{z}}^{n}}^{*}+{\bar{a}}_0b_{\overline{u\varphi }}^{*}S_{z^n}&={\bar{a}}_0t_{z^{n}}b_{\overline{u\varphi }}^{*}+b_{{\bar{z}}^{n}}^{*}S_{\varphi }, \end{aligned}$$

(11)

$$\begin{aligned} b_{{\bar{u}}\varphi }t_{z^{n}}&=S_{z^{n}}b_{{\bar{u}}\varphi }, \end{aligned}$$

(12)

$$\begin{aligned} \frac{1}{{\bar{a}}_0}b_{{\bar{u}}\varphi }b_{{\bar{z}}^{n}}^{*}+S_{\varphi }S_{z^{n}}&=S_{z^{n}}S_{\varphi }. \end{aligned}$$

(13)

For any $x \in H^2, ux \in uH^2$, we have

$$\begin{aligned} t_{\varphi }t_{z^{n}}ux-t_{z^{n}}t_{\varphi }ux&=t_{\varphi }uP{\bar{u}}z^{n}ux-t_{z^{n}}uP{\bar{u}}\varphi ux\\&=uP{\bar{u}}\varphi uz^{n}x-uP{\bar{u}}z^{n}uP\varphi x\\&=uP(\varphi z^{n}x)-uP(z^nP\varphi x),\\ \frac{1}{{\bar{a}}_0}b_{{\bar{z}}^{n}}^{*}b_{{\bar{u}}\varphi }ux&=\frac{1}{{\bar{a}}_0}b_{{\bar{z}}^{n}}^{*}P_{-}({\bar{u}}\varphi ux)\\&=\frac{1}{{\bar{a}}_0}uP({\bar{u}}z^nP_{-}(\varphi x)). \end{aligned}$$

Suppose $a_s\; (s \ge 1)$ is the second nonzero coefficent of u, that is, $a_0 \ne 0, \; a_1=a_2=\dots =a_{s-1}=0,$ and $a_s \ne 0$. If we take $n=s+1$ and $x=1$, then

$$\begin{aligned} t_{\varphi }t_{z^{s+1}} u-t_{z^{s+1}}t_{\varphi }u =&uP(\varphi z^{s+1})-uP(z^{s+1}P\varphi )\\ =&uz^{s+1}\left(\sum \limits _{m=0}^{+ \infty } c_{m}z^m+ \sum \limits _{m=1}^{s+1} c_{-m}{\bar{z}}^m\right)-uz^{s+1}\sum \limits _{m=0}^{+ \infty } c_{m}z^m\\ =&uz^{s+1}\sum \limits _{m=1}^{s+1} c_{-m}{\bar{z}}^m\\ =&u(c_{-(s+1)}+c_{-s}z+c_{-s+1}z^2+\dots +c_{-1}z^s),\\ \frac{1}{{\bar{a}}_0}b_{{\bar{z}}^{s+1}}^{*}b_{{\bar{u}}\varphi }u =&\frac{1}{{\bar{a}}_0}uP({\bar{u}}z^{s+1}P_{-}(\varphi ))\\ =&\frac{1}{{\bar{a}}_0}uP(z^{s+1}({\bar{a}}_0+\sum \limits _{n=s}^{+ \infty } {\bar{a}}_{n}{\bar{z}}^n) \sum \limits _{m=1}^{+ \infty } c_{-m}{\bar{z}}^m)\\ =&\frac{1}{{\bar{a}}_0}u({\bar{a}}_0z^{s+1}\sum \limits _{m=1}^{s+1} c_{-m}{\bar{z}}^m+{\bar{a}}_sc_{-1})\\ =&\frac{1}{{\bar{a}}_0}u\left[({\bar{a}}_0c_{-(s+1)}+{\bar{a}}_sc_{-1})+{\bar{a}}_0c_{-s}z+{\bar{a}}_0c_{-s+1}z^2+\dots \right. \\&\left. + {\bar{a}}_0c_{-1}z^s\right]. \end{aligned}$$

Since u is nonconstant, by equality (10), the above two identities give that

$$\begin{aligned} {\bar{a}}_0(c_{-(s+1)}+c_{-s}z+c_{-s+1}z^2+\dots +c_{-1}z^s)&=({\bar{a}}_0c_{-(s+1)}+{\bar{a}}_sc_{-1})+{\bar{a}}_0c_{-s}z\\&+{\bar{a}}_0c_{-s+1}z^2+\dots + {\bar{a}}_0c_{-1}z^s. \end{aligned}$$

Comparing the constant terms on both sides, we get $c_{-1}=0$ since $a_s \ne 0$. Fix $n=s+1$ and repeat the above process for $x=z^i$ for any positive integer i; by equality (10), a simple computation gives that $c_{-(m+1)}=0$ for all $m \geqslant 1.$ It follows that $\varphi \in H^{\infty }$. Furthermore, by equation (11), we have

$$\begin{aligned} t_{\varphi }b_{{\bar{z}}^{n}}^{*}-b_{{\bar{z}}^{n}}^{*}S_{\varphi }={\bar{a}}_0(t_{z^{n}}b_{\overline{u\varphi }}^{*}-b_{\overline{u\varphi }}^{*}S_{z^n}). \end{aligned}$$

An easy computation gives that

$$\begin{aligned} (t_{\varphi }b_{{\bar{z}}^{s+1}}^{*}-b_{{\bar{z}}^{s+1}}^{*}S_{\varphi }){\bar{z}}&=uP{\bar{u}}\varphi uP{\bar{u}}z^{s+1}{\bar{z}}-uP{\bar{u}}z^{s+1}P_{-}\varphi {\bar{z}}\\&=uP\varphi P{\bar{u}}z^s-uP{\bar{u}}z^{s+1}P_{-}\varphi {\bar{z}}\\&=u\varphi ({\bar{a}}_0z^s+{\bar{a}}_s)-c_0u({\bar{a}}_0z^s+{\bar{a}}_s)\\&=u({\bar{a}}_0z^s+{\bar{a}}_s)\sum _{m=1}^{\infty }c_mz^m\\&={\bar{a}}_0u\sum _{m=1}^{\infty }c_mz^{m+s}+{\bar{a}}_su\sum _{m=1}^{\infty }c_mz^{m},\\ {\bar{a}}_0(t_{z^{s+1}}b_{\overline{u\varphi }}^{*}-b_{\overline{u\varphi }}^{*}S_{z^{s+1}}){\bar{z}}&={\bar{a}}_0uP{\bar{u}}z^{s+1}uP{\bar{u}}u\varphi {\bar{z}}-{\bar{a}}_0uP{\bar{u}}u\varphi P_{-}(z^{s+1}{\bar{z}})\\&={\bar{a}}_0uPz^{s+1}P(\varphi {\bar{z}})\\&={\bar{a}}_0uz^{s+1}\sum _{m=1}^{\infty }c_mz^{m-1}\\&={\bar{a}}_0u\sum _{m=1}^{\infty }c_mz^{m+s}; \end{aligned}$$

hence we obtain ${\bar{a}}_su\sum \nolimits _{m=1}^{\infty }c_mz^{m}=0$, which gives that ${\bar{a}}_sc_m=0$ for all $m \geqslant 1$ since u is nonconstant. Because $a_s \ne 0$, it follows that $c_m=0$ for all $m \geqslant 1$; hence $\varphi$ is a constant. Moreover, $b_{{\bar{u}}\varphi }ux =0$ for any $x \in H^2$; thus $b_{{\bar{u}}\varphi } =0$. Hence $T=c_0I$, and this proves (9) for the case $k = 0$. Therefore, the result follows.

(b) For the case $k \geqslant 1$, that is $a_0 =a_1=\dots =a_{k-1}=0$ and $a_k \ne 0$, then $u(z)=z^k\sum \nolimits _{m=0}^{+\infty }a_{m+k}z^m$. Because $TD_z=D_zT$, Theorem 1 gives that there exist some $\varphi , \psi \in H^{\infty }$ and $f \in L^{\infty }$ such that

$$\begin{aligned} T=\left[ \begin{array}{ccccc} t_{\varphi } &{} 0 \\ b_{{\bar{u}}f} &{} S_{\psi } \end{array}\right] . \end{aligned}$$

Since

$$\begin{aligned} D_{z^{n}}=\left[ \begin{array}{ccccc} t_{z^{n}}&{} b_{{\bar{z}}^{n}}^{*}\\ 0&{} S_{z^{n}} \end{array}\right] , \end{aligned}$$

and $TD_{z^{n}}=D_{z^{n}}T \; (n \geqslant 1)$, it follows that

$$\begin{aligned} \left[ \begin{array}{ccccc} t_{\varphi } &{} 0 \\ b_{{\bar{u}}f} &{} S_{\psi } \end{array}\right] \left[ \begin{array}{ccccc} t_{z^{n}}&{} b_{{\bar{z}}^{n}}^{*}\\ 0&{} S_{z^{n}} \end{array}\right] =\left[ \begin{array}{ccccc} t_{z^{n}}&{} b_{{\bar{z}}^{n}}^{*}\\ 0&{} S_{z^{n}} \end{array}\right] \left[ \begin{array}{ccccc} t_{\varphi } &{} 0 \\ b_{{\bar{u}}f} &{} S_{\psi } \end{array}\right] . \end{aligned}$$

Hence we get the following four equations:

$$\begin{aligned} t_{\varphi }t_{z^{n}}&=t_{z^{n}}t_{\varphi }+b_{{\bar{z}}^{n}}^{*}b_{{\bar{u}}f}, \end{aligned}$$

(14)

$$\begin{aligned} t_{\varphi }b_{{\bar{z}}^{n}}^{*}&=b_{{\bar{z}}^{n}}^{*}S_{\psi }, \end{aligned}$$

(15)

$$\begin{aligned} b_{{\bar{u}}f}t_{z^{n}}&=S_{z^{n}}b_{{\bar{u}}f}, \end{aligned}$$

(16)

$$\begin{aligned} b_{{\bar{u}}f}b_{{\bar{z}}^{n}}^{*}+S_{\psi }S_{z^{n}}&=S_{z^{n}}S_{\psi }. \end{aligned}$$

(17)

For $\varphi \in H^{\infty }$, since $t_{\varphi }t_{z^{n}}=t_{\varphi z^{n}}=t_{z^{n}}t_{\psi }$, Eq. (14) gives that

$$\begin{aligned} b_{{\bar{z}}^{n}}^{*}b_{{\bar{u}}f}=0. \end{aligned}$$

(18)

Thus for any $x \in H^2$, $ux \in uH^2$, we have

$$\begin{aligned} 0=b_{{\bar{z}}^{n}}^{*}b_{{\bar{u}}f}ux=b_{{\bar{z}}^{n}}^{*}(P_{-}({\bar{u}}fux)=b_{{\bar{z}}^{n}}^{*}(P_{-}(fx)) =uP{\bar{u}}z^{n}(P_{-}(fx)). \end{aligned}$$

(19)

Suppose $f \in L^{\infty }$ with Fourier series $f(z)=\sum \nolimits _{m=1}^{+ \infty } l_{-m}{\bar{z}}^m+\sum \nolimits _{m=0}^{+ \infty } l_mz^m$, then $P_{-}(fz^i)=\sum \nolimits _{m=i+1}^{+ \infty } l_{-m}{\bar{z}}^{m-i}$ for any $i \in {\mathbb {N}}$. If we take $n=k+1$ and $x=z^i \; (i \ge 0)$, it follows from Eq.  (19) that

$$\begin{aligned} 0=uP{\bar{u}}z^{k+1}(P_{-}(fz^{i}))=u{\bar{a}}_kl_{-(i+1)}. \end{aligned}$$

Since $a_k \ne 0$ and u is nonconstant, we have $l_{-(i+1)}=0$ for any $i \in {\mathbb {N}}$. Therefore, f is analytic. It follows that $b_{{\bar{u}}f}=0$ since $b_{{\bar{u}}f}ux=P_{-}(fx)=0$ for any $x \in H^2$.

In the following, we will use Eq. (15) to obtain that $\varphi =\psi =c$ with c constant. For $\varphi , \psi \in H^{\infty }$, suppose $\varphi =\sum \nolimits _{m=0}^{+ \infty } c_mz^m$ and $\psi =\sum \nolimits _{m=0}^{+ \infty } d_mz^m$. Then for any $y \in (H^2)^{\bot }$, we obtain

$$\begin{aligned} t_{\varphi }b_{{\bar{z}}^{n}}^{*}y=uP({\bar{u}}\varphi (uP{\bar{u}}{z}^{n}y))=uP(\varphi P{\bar{u}}{z}^{n}y)=u\varphi P{\bar{u}}{z}^{n}y, \end{aligned}$$

(20)

$$\begin{aligned} b_{{\bar{z}}^{n}}^{*}S_{\psi }y=b_{{\bar{z}}^{n}}^{*}P_{-}(\psi y)=uP({\bar{u}}{z}^{n}P_{-}(\psi y)). \end{aligned}$$

(21)

We will use (15) applied to $y={\bar{z}}^j$ in the following. Note that $P({\bar{u}}{z}^{n}{\bar{z}}^j)=\sum \nolimits _{m=k}^{n-j}{\bar{a}}_mz^{(n-j)-m}$ for $j < n$, then we have

$$\begin{aligned} t_{\varphi }b_{{\bar{z}}^{n}}^{*}{\bar{z}}^{j}=u\varphi \sum \limits _{m=k}^{n-j}{\bar{a}}_mz^{(n-j)-m}. \end{aligned}$$

Moreover, a direct calculation gives that $P_{-}(\psi {\bar{z}}^j)=\sum \nolimits _{m=0}^{j-1} d_m{\bar{z}}^{j-m}$. If we take $n=k+1, j=1$, we have $t_{\varphi }b_{{\bar{z}}^{k+1}}^{*}{\bar{z}}={\bar{a}}_ku\varphi$ and $b_{{\bar{z}}^{k+1}}^{*}S_{\psi }{\bar{z}}={\bar{a}}_kd_0u$; then by Eq. (15) and $a_k \ne 0$, we obtain $\varphi = d_0$, i.e. $c_0=d_0$ and $c_m=0$ for all $m \geqslant 1$. If we take $n=k+2, j=2$, then equalities (20) and (21) give that $t_{\varphi }b_{{\bar{z}}^{k+2}}^{*}{\bar{z}}^{2}={\bar{a}}_ku\varphi ={\bar{a}}_kc_0u$ and $b_{{\bar{z}}^{k+2}}^{*}S_{\psi }{\bar{z}}^{2}=u[({\bar{a}}_kd_0+{\bar{a}}_{k+1}d_1)+{\bar{a}}_kd_1z]=u[({\bar{a}}_kc_0+{\bar{a}}_{k+1}d_1)+{\bar{a}}_kd_1z]$, respectively. Using Eq. (15) again, and comparing the coefficients, we have ${\bar{a}}_kd_1=0$; thus $d_1=0$ because $a_k \ne 0$. Lemma 2 implies the following result for (20) and (21) by taking $n=k+j$ and $y={\bar{z}}^j$ for any $j \ge 2$, we get $d_m=0$ for all $m \geqslant 1$, it follows that $\psi =d_0$. Hence, $\varphi =\psi = c_0$ is constant. So we have

$$\begin{aligned} T=cI, \, c \in {\mathbb {C}} \end{aligned}$$

if $T \in \{D_{z^n}: n \in {\mathbb {N}}\}^{'}$. Therefore, this proves the result for the case $k \ge 1$. $\square$

For $g \in H^{\infty }$ with Fourier series $g(z)=\sum \nolimits _{n=0}^{\infty }g_nz^n$, by the linearity of dual truncated Toeplitz operator $D_g$ with respect to its symbol, we obtain the following result:

Theorem 2

Let u be a nonconstant inner function. Then

$$\begin{aligned} \{D_g: g \in H^{\infty }\}^{'}=\left\{ cI: \, c \in {\mathbb {C}} \right\} . \end{aligned}$$

4 The invariant subspaces of $D_z$

In this section, we mainly seek the structure of nontrivial invariant subspaces of $D_z$ by applying the methods and techniques on classical Hardy spaces.

Theorem 3

Let u be a nonconstant inner function with $u(0)=0$ and ${\mathcal {M}}$ be a nontrivial closed subspace of $(K_u^2)^{\bot }$. Then one of the following cases happens:

• (a) If ${\mathcal {M}} \subset uH^2$, then ${\mathcal {M}}$ is an invariant subspace of $D_z$ if and only if there exists a nonconstant inner function $\theta$ such that

  $$\begin{aligned} {\mathcal {M}}=\theta uH^2; \end{aligned}$$

• (b) If ${\mathcal {M}} \subset (H^2)^{\bot }$, then ${\mathcal {M}}$ is an invariant subspace of $D_z$ if and only if there exists a nonconstant inner function $\eta$ such that

  $$\begin{aligned} {\mathcal {M}}={\bar{z}} \overline{K_{\eta }^2}; \end{aligned}$$

• (c) Let ${\mathcal {M}} \nsubseteq uH^2$ and ${\mathcal {M}} \nsubseteq (H^2)^{\bot }$. If ${\mathcal {M}}$ is an invariant subspace of $D_z$, then there exist nonconstant inner functions $\theta$ and $\eta$ such that $cl(uP{\bar{u}}{\mathcal {M}})=\theta uH^2$ and $cl(P_{-}{\mathcal {M}})={\bar{z}} \overline{K_{\eta }^2},$ where $cl({\mathcal {N}})$ denotes the closure of ${\mathcal {N}}$. Conversely, if there exist inner functions $\theta$ and $\eta$ with one of them nonconstant such that ${\mathcal {M}}_{+}=cl(uP{\bar{u}}{\mathcal {M}})=\theta uH^2$ and ${\mathcal {M}}_{-}=cl(P_{-}{\mathcal {M}})={\bar{z}} \overline{K_{\eta }^2}$, then ${\mathcal {M}}_{+} \oplus {\mathcal {M}}_{-}$ is a nontrivial invariant subspace of $D_z$.

Proof

Since $u(0)=0$, $D_z$ has a matrix decomposition on $(K_u^2)^{\bot }= uH^2 \oplus (H^2)^{\bot }$ as follows:

$$\begin{aligned} D_z=\left[ \begin{array}{ccccc} t_{z}&{}0\\ 0&{}S_{z} \end{array}\right] . \end{aligned}$$

Suppose ${\mathcal {M}}$ is an invariant subspace of $D_z$, that is, $D_z{\mathcal {M}} \subset {\mathcal {M}}$.

(a) If ${\mathcal {M}} \subset uH^2$, then we have $t_z{\mathcal {M}} \subset {\mathcal {M}}$. Because $t_z=M_uT_zM_{{\bar{u}}}$, we get $M_uT_zM_{{\bar{u}}}{\mathcal {M}} \subset {\mathcal {M}}$; thus $T_zM_{{\bar{u}}}{\mathcal {M}} \subset M_{{\bar{u}}}{\mathcal {M}}={\bar{u}}{\mathcal {M}}.$ By the famous Beurling’s theorem [2], there is an inner function $\theta$ such that ${\bar{u}}{\mathcal {M}}=\theta H^2$; hence ${\mathcal {M}}=\theta uH^2$.

(b) If ${\mathcal {M}} \subset (H^2)^{\bot }$, then we have $S_z{\mathcal {M}} \subset {\mathcal {M}}$. Since $S_z=VT_{{\bar{z}}}V$, we obtain $VT_{{\bar{z}}}V{\mathcal {M}} \subset {\mathcal {M}}$; thus $T_{{\bar{z}}}V{\mathcal {M}} \subset V{\mathcal {M}}.$ Using Beurling’s theorem again, there exists some inner function $\eta$ such that $V{\mathcal {M}}=H^2 \ominus \eta H^2=K_{\eta }^2$. It follows that ${\mathcal {M}}=VK_{\eta }^2={\bar{z}}\overline{K_{\eta }^2}$.

(c) If ${\mathcal {M}} \nsubseteq uH^2$ and ${\mathcal {M}} \nsubseteq (H^2)^{\bot }$, let ${\mathcal {M}}_{1}=uP{\bar{u}}{\mathcal {M}}=\{uP{\bar{u}}x: x \in {\mathcal {M}}\}$ and ${\mathcal {M}}_{2}=P_-{\mathcal {M}}=\{P_{-}x: x \in {\mathcal {M}}\},$ then ${\mathcal {M}} \subset {\mathcal {M}}_1 + {\mathcal {M}}_{2}$. For any $x \in {\mathcal {M}}$ and $x= x_{1} +x_{2}$ with $x_{1} \in {\mathcal {M}}_{1}$ and $x_{2} \in {\mathcal {M}}_{2}$, we have

$$\begin{aligned} D_zx=\left[ \begin{array}{ccccc} t_{z}&{}0\\ 0&{}S_{z} \end{array}\right] \left[ \begin{array}{ccccc} x_{1}\\ x_{2} \end{array}\right] =\left[ \begin{array}{ccccc} t_{z}x_{1}\\ S_{z}x_{2} \end{array}\right] \in {\mathcal {M}}. \end{aligned}$$

Hence $t_z{\mathcal {M}}_1 \subset {\mathcal {M}}_1, \, S_z{\mathcal {M}}_2 \subset {\mathcal {M}}_2;$ it follows that $t_zcl({\mathcal {M}}_1) \subset cl({\mathcal {M}}_1),$ $S_zcl({\mathcal {M}}_2)$ $\subset cl({\mathcal {M}}_2).$ By (a) and (b), there exists some inner functions $\theta$ and $\eta$ such that $cl(uP{\bar{u}}{\mathcal {M}})=cl({\mathcal {M}}_1)=\theta uH^2$ and $cl(P_{-}{\mathcal {M}})=cl({\mathcal {M}}_2)={\bar{z}}\overline{K_{\eta }^2}$. The converse direction is obvious. Therefore, this completes the proof. $\square$

The following examples show that the above characterization of invariant subspaces for $D_z$ with inner function u satisfying $u(0)=0$ is the best we can obtain. The first example tells us that for any nontrivial invariant subspace ${\mathcal {M}}$ for $D_z$, in general, ${\mathcal {M}}$ does not have a decomposition as ${\mathcal {M}} = {\mathcal {M}}_{+} \oplus {\mathcal {M}}_{-}$ with ${\mathcal {M}}_+$ being a closed subspace of $uH^2$ and ${\mathcal {M}}_-$ being a closed subspace of $\overline{zH^2}$.

Example 1

Suppose u and $\varphi$ are nonconstant inner functions with $u(0)=0$, and $\psi =\sum \nolimits _{j=1}^{\infty }a_jz^j \in H^{\infty }$ with $a_j \ne 0$ for all $j \in {\mathbb {N}}$. Let $x=zu\varphi +{\bar{\psi }}$; then $x \in uH^2+\overline{zH^2}$. Let ${\mathcal {M}}$ be the closed linear subspace generated by $\{x, D_zx, D_z^2x, \dots , D_z^nx, \dots \}$, and, let ${\mathcal {M}}_{+}$ and ${\mathcal {M}}_{-}$ be the closed subspaces generated by $\{uz^{n+1}\varphi \}$ and $\{P_{-}z^n{\bar{\psi }}\}_{n \ge 0}$, respectively. Then ${\mathcal {M}}$ is a nontrivial invariant subspace for $D_z$, but ${\mathcal {M}} \ne {\mathcal {M}}_{+} \oplus {\mathcal {M}}_{-}$.

Proof

Obviously, ${\mathcal {M}}$ is an invariant subspace for $D_z$ and ${\mathcal {M}}$ is nontrivial since $u \perp {\mathcal {M}}$. It is clear that ${\mathcal {M}} \subset {\mathcal {M}}_{+} \oplus {\mathcal {M}}_{-}$. In the following, we will show that conjugate analytic functions are not included in ${\mathcal {M}}$, which implies that ${\mathcal {M}} \subsetneqq {\mathcal {M}}_{+} \oplus {\mathcal {M}}_{-}$. For any analytic polynomial ${\mathcal {P}}_k(z)$, we get

$$\begin{aligned} \Vert {\mathcal {P}}_k(D_z)x\Vert ^2_2=\Vert zu\varphi {\mathcal {P}}_k(z)+S_{{\mathcal {P}}_k(z)}{\bar{\psi }}\Vert ^2_2 =\Vert zu\varphi {\mathcal {P}}_k(z)\Vert ^2_2+\Vert P_{-}({\mathcal {P}}_k(z){\bar{\psi }})\Vert ^2_2. \end{aligned}$$

Hence $\{{\mathcal {P}}_k(D_z)x\}$ is convergent in $L^2$-norm if and only if both $\{zu\varphi {\mathcal {P}}_k(z)\}$ and $\{P_{-}({\mathcal {P}}_k(z){\bar{\psi }})\}$ are convergent in $L^2$-norm. If $zu\varphi {\mathcal {P}}_k(z)$ converges to 0 in $L^2$-norm, then

$$\begin{aligned} \Vert zu\varphi {\mathcal {P}}_k(z)\Vert _2 = \Vert {\mathcal {P}}_k(z)\Vert _2 \rightarrow 0 \;\;\text {as} \;\; k \rightarrow \infty , \end{aligned}$$

it follows that

$$\begin{aligned} \Vert P_{-}({\mathcal {P}}_k(z){\bar{\psi }})\Vert \le \Vert {\mathcal {P}}_k(z){\bar{\psi }}\Vert _2 \le \Vert {\bar{\psi }}\Vert _{\infty }\Vert {\mathcal {P}}_k(z)\Vert _2 \rightarrow 0 \;\;\text {as} \;\; k \rightarrow \infty . \end{aligned}$$

Thus there is no conjugate analytic functions in ${\mathcal {M}}$. Therefore, we arrive at the assertion. $\square$

For any closed subspace ${\mathcal {M}} \subset (K_u^2)^{\bot }$ with $u(0)=0$, suppose ${\mathcal {M}}_{+}$ and ${\mathcal {M}}_{-}$ are the closed subspaces generated by $uP{\bar{u}}{\mathcal {M}}$ and $P_{-}{\mathcal {M}}$, respectively. If ${\mathcal {M}}_{+} \oplus {\mathcal {M}}_{-}$ is a nontrivial invariant subspace for $D_z$, do we have $D_z{\mathcal {M}} \subset {\mathcal {M}}?$ Actually, the example we construct below says that the answer is negative.

Example 2

Suppose u and $\varphi$ are nonconstant inner functions with $u(0)=0$, and $\psi =\sum \nolimits _{j=1}^{\infty }\frac{1}{j^2}z^j$. Let $x=zu\varphi +{\bar{\psi }}$ and let ${\mathcal {M}}$ be the closed subspace linearly generated by $\{{\bar{\psi }}, x, D_zx, D_z^2x, \dots , D_z^nx, \dots \}$, ${\mathcal {M}}_{+}$ and ${\mathcal {M}}_{-}$ be the closed subspace generated by $\{uz^{n+1}\varphi \}_{n \ge 0}$ and $\{P_{-}z^n{\bar{\psi }}\}_{n \ge 0}$, respectively. Then ${\mathcal {M}}_{+} \oplus {\mathcal {M}}_{-}$ is an invariant subspace for $D_z$, but ${\mathcal {M}}$ is not an invariant subspace for $D_z$.

Proof

It is obvious that ${\mathcal {M}}_{+} \oplus {\mathcal {M}}_{-}$ is an invariant subspace for $D_z$. We only need to show that ${\mathcal {M}}$ is not an invariant subspace for $D_z$. Since ${\bar{\psi }} \in {\mathcal {M}}$ and $x \in {\mathcal {M}}$, we have $zu\varphi \in {\mathcal {M}}$. For any $y \in {\mathcal {M}}$, y is approximated by $A_{k}zu\varphi +B_{k}{\bar{\psi }}+{\mathcal {P}}_{k}(D_z)x$ in $L^2$-norm, where $A_{k}, B_{k}$ are constant and ${\mathcal {P}}_{k}(z)$ is an analytic polynomial, and

$$\begin{aligned} A_{k}zu\varphi +B_{k}{\bar{\psi }}+{\mathcal {P}}_{k}(D_z)x= A_{k}zu\varphi +zu\varphi {\mathcal {P}}_k(z)+B_{k}{\bar{\psi }}+P_{-}({\mathcal {P}}_k(z){\bar{\psi }}). \end{aligned}$$

Then we have

$$\begin{aligned} \Vert A_{k}zu\varphi +B_{k}{\bar{\psi }}+{\mathcal {P}}_{k}(D_z)x\Vert _2^2&= \Vert A_{k}zu\varphi +zu\varphi {\mathcal {P}}_k(z)\Vert _2^2+\Vert B_{k}{\bar{\psi }}+P_{-}({\mathcal {P}}_k(z){\bar{\psi }})\Vert _2^2\\&=\Vert zu\varphi (A_{k}+{\mathcal {P}}_k(z)) \Vert _2^2+\Vert P_{-}(B_{k}+{\mathcal {P}}_k(z)){\bar{\psi }}\Vert _2^2\\&=\Vert A_{k}+{\mathcal {P}}_k(z) \Vert _2^2+\Vert P_{-}(B_{k}+{\mathcal {P}}_k(z)){\bar{\psi }}\Vert _2^2. \end{aligned}$$

If $\Vert A_{k}zu\varphi +B_{k}{\bar{\psi }}+{\mathcal {P}}_{k}(D_z)x\Vert$ is convergent, then $\Vert zu\varphi (A_{k}+{\mathcal {P}}_k(z)) \Vert _2 \rightarrow 0$ as $k \rightarrow +\infty$ if and only if $\Vert A_{k}+{\mathcal {P}}_k(z) \Vert _2 \rightarrow 0$. Since

$$\begin{aligned} \Vert P_{-}(B_{k}+{\mathcal {P}}_k(z)){\bar{\psi }}\Vert _2&=\Vert P_{-}(B_{k}-A_k){\bar{\psi }} +P_{-}(A_{k}+{\mathcal {P}}_k(z)){\bar{\psi }}\Vert _2 \end{aligned}$$

and $P_{-}(A_{k}+{\mathcal {P}}_k(z) )$ tends to zero, we get

$$\begin{aligned} \lim \limits _{k \rightarrow + \infty } \Vert P_{-}(B_{k}+{\mathcal {P}}_k(z)){\bar{\psi }}\Vert _2&=\lim \limits _{k \rightarrow + \infty } \Vert P_{-}(B_{k}-A_k){\bar{\psi }}\Vert _2\\&=\lim \limits _{k \rightarrow + \infty } |B_{k}-A_k|\Vert {\bar{\psi }}\Vert _2, \end{aligned}$$

which means that $P_{-}(B_{k}+{\mathcal {P}}_k(z)){\bar{\psi }}$ converges to $c{\bar{\psi }}$ with c constant, it follows that the conjugate functions in ${\mathcal {M}}$ are of form $c{\bar{\psi }}$ with c constant. So $D_z^n{\bar{\psi }} =P_{-}z^n{\bar{\psi }} \notin {\mathcal {M}}$, which implies that ${\mathcal {M}}$ is not an invariant subspace for $D_z$. $\square$

If u is a nonconstant inner function with $u(0) \ne 0$, then the structure of invariant subspaces for $D_z$ is described as follows:

Theorem 4

Let u be a nonconstant inner function with $u(0) \ne 0$. Then the nontrivial closed subspace ${\mathcal {M}}$ of $(K_u^2)^{\bot }$ is an invariant subspace of $D_z$ if and only if one of the following cases happens:

• (a) there exists a unimodular function $\varphi$ such that

  $$\begin{aligned} {\mathcal {M}}=U_A( \varphi H^2), \end{aligned}$$

  where $U_A=\left[ \begin{array}{ccccc} M_u&{}0\\ 0&{}\frac{1}{{\bar{u}}(0)}I_{\overline{zH^2}} \end{array}\right]$ on $L^2=H^2 \oplus \overline{zH^2}$;

• (b) there exists a Borel set $E \subset {\mathbb {T}}$ with $0< m_E <1$ such that

  $$\begin{aligned} {\mathcal {M}}=U_AL_{E}^2, \end{aligned}$$

  where $m_E$ is the Lebesgue measure of E and $L_E^2=\{f \in L^2: f|_{{\mathbb {T}}-E}=0\}$.

Proof

Since $u(0) \ne 0$, $D_z$ has a matrix representation on $(K_u^2)^{\bot }= uH^2 \oplus \overline{zH^2}$ as follows:

$$\begin{aligned} D_z=\left[ \begin{array}{ccccc} t_{z}&{}b_{{\bar{z}}}^{*}\\ 0&{}S_{z} \end{array}\right] . \end{aligned}$$

On the other hand, on $L^2=H^2 \oplus \overline{zH^{2}}$, suppose

$$\begin{aligned} A=\left[ \begin{array}{ccccc} I_{H_2}&{}0\\ 0&{}\frac{1}{{\bar{u}}(0)}I_{\overline{zH_2}} \end{array}\right] ; \end{aligned}$$

then

$$\begin{aligned} A^{-1}=\left[ \begin{array}{ccccc} I_{H_2}&{}0\\ 0&{}{\bar{u}}(0)I_{\overline{zH_2}} \end{array}\right] . \end{aligned}$$

By the proof of Theorem 1 in Sect. 3, we have $AM_z={\widetilde{D}}_zA$; thus ${\widetilde{D}}_z=AM_zA^{-1}$. Suppose ${\mathcal {M}} \subset (K_u^2)^{\bot }$ is a nontrivial invariant subspace of $D_z$, i.e. $D_z{\mathcal {M}} \subset {\mathcal {M}}.$ Note that $D_z=U{\widetilde{D}}_zU^{*}$; then we get $U{\widetilde{D}}_zU^{*}{\mathcal {M}}\subset {\mathcal {M}}$, which implies that ${\widetilde{D}}_zU^{*}{\mathcal {M}}\subset U^{*}{\mathcal {M}};$ here U is a unitary operator on $L^2=H^2 \oplus \overline{zH^{2}}$ defined by

$$\begin{aligned} U=\left[ \begin{array}{ccccc} M_u&{}0\\ 0&{}I_{\overline{zH_2}} \end{array}\right] . \end{aligned}$$

Let $\widetilde{{\mathcal {M}}}=U^{*}{\mathcal {M}}$; then we have ${\widetilde{D}}_z\widetilde{{\mathcal {M}}} \subset \widetilde{{\mathcal {M}}}$; it follows that

$$\begin{aligned} AM_zA^{-1}\widetilde{{\mathcal {M}}}\subset \widetilde{{\mathcal {M}}}, and \end{aligned}$$

hence $M_zA^{-1}\widetilde{{\mathcal {M}}}\subset A^{-1}\widetilde{{\mathcal {M}}}$, that is, $A^{-1}\widetilde{{\mathcal {M}}}$ is an invariant subspace of $M_z$. Then $A^{-1}\widetilde{{\mathcal {M}}}$ has a unique direct sum decomposition $A^{-1}\widetilde{{\mathcal {M}}}=\widetilde{{\mathcal {N}}_1} \oplus \widetilde{{\mathcal {N}}_2}$ by ([10, Theorem 6.12]) such that $z\widetilde{{\mathcal {N}}_1}=\widetilde{{\mathcal {N}}_1}, \bigcap \limits _{n \geqslant 0}z^n\widetilde{{\mathcal {N}}_2}=\{0\}$ and both ${\mathcal {N}}_1$ and ${\mathcal {N}}_2$ are invariant subspaces of $M_z$. By [14, Part A,Theorem 1.2.1], there exists a Borel set $E \subset {\mathbb {T}}$ such that

$$\begin{aligned} \widetilde{{\mathcal {N}}_1}=L_{E}^2=\{f \in L^2: f(z)=0, z \notin E\}. \end{aligned}$$

Now it maybe happen that $\widetilde{{\mathcal {N}}_2}=\{0\}$ or $\widetilde{{\mathcal {N}}_2} \ne \{0\}$. If $\widetilde{{\mathcal {N}}_2}\ne \{0\}$, then by Beurling’s Theorem ([14, Part A, Corollary 1.4.1]), there exists a unimodular function $\varphi$ such that $\widetilde{{\mathcal {N}}_2}=\varphi H^2.$ Note that $\varphi L_{E}^2 = L_{E}^2$, and $L_{E}^2 \perp \varphi H^2$; thus $\varphi L_{E}^2 \perp \varphi H^2$. For any $f \in L_{E}^2$, since $\varphi f \perp \varphi H^2$, we have $f \perp H^2$, it follows that ${\bar{f}}$ is analytic. Further, since $A^{-1}\widetilde{{\mathcal {M}}}$ is a nontrivial invariant subspace of $M_z$, that is, $\widetilde{{\mathcal {M}}} \ne L^2$, we have $L_{E}^2 \ne L^2$; thus the Lebesgue measure of E is strictly less than 1. Moreover, because $f|_{{\mathbb {T}}-E}=0$, we get ${\bar{f}}=0$ a.e. on ${\mathbb {T}}$ by Corollary 1.4.3 in [14], it follows that $f=0$ a.e. on ${\mathbb {T}}$ . Hence $L_{E}^2=\{0\}$, and thus $A^{-1}\widetilde{{\mathcal {M}}}=\widetilde{{\mathcal {N}}_2}=\varphi H^2.$ Since $\widetilde{{\mathcal {M}}}=U^{*}{\mathcal {M}}$, we have

$$\begin{aligned} {\mathcal {M}}=U\widetilde{{\mathcal {M}}}=UA\varphi H^2. \end{aligned}$$

The remaining case happens if $\widetilde{{\mathcal {N}}_2}=0$, then $A^{-1}\widetilde{{\mathcal {M}}}=\widetilde{{\mathcal {N}}_1}=L_{E}^2;$ it follows that

$$\begin{aligned} {\mathcal {M}}=UA\widetilde{{\mathcal {M}}}=UAL_{E}^2, \end{aligned}$$

where E is a Borel subset of ${\mathbb {T}}$ with $0< m_E <1$. Let $U_A=UA$; therefore, the result is obtained. $\square$

Recall that for $T \in B({\mathcal {H}})$, a closed subspace ${\mathcal {M}}$ is said to be a reducing subspace for T if $T{\mathcal {M}} \subset {\mathcal {M}}$ and $T{\mathcal {M}}^{\bot } \subset {\mathcal {M}}^{\bot }$; then by ([10, Proposition  4.42]), it is equivalent to say that ${\mathcal {M}}$ is a reducing subspace for T if and only if $TP_{{\mathcal {M}}}=P_{{\mathcal {M}}}T$, where $P_{{\mathcal {M}}}$ is the orthogonal projection from ${\mathcal {H}}$ to ${\mathcal {M}}$. It follows that $P_{{\mathcal {M}}} \in \{T\}^{'}$ if ${\mathcal {M}}$ is a reducing subspace for T.

Combining the above fact of reducing subspace with Theorem 1, the following result is naturally obtained. We also want to point out that the case $u(0) \ne 0$ was mentioned in [13] without a detailed proof but the case $u(0) = 0$ was not included in [13]. For the reader’s convenience, we present a different proof here.

Corollary 1

Let u be a nonconstant inner function.

(1) If $u(0) = 0$, then the nontrivial reducing subspace for $D_z$ is either $uH^2$ or $\overline{zH^{2}}$.

(2) If $u(0) \ne 0$, then $D_z$ has no nontrivial reducing subspace.

Proof

Assume ${\mathcal {M}}$ is a reducing subspace of $D_z$; then we have

$$\begin{aligned} D_zP_{{\mathcal {M}}}=P_{{\mathcal {M}}}D_z, \end{aligned}$$

where $P_{{\mathcal {M}}}$ is the orthogonal projection from $(K_u^2)^{\bot }$ to ${\mathcal {M}}$. This implies that $P_{{\mathcal {M}}} \in \{D_z\}^{'}$.

(1) If $u(0) = 0$, then Theorem 1 gives that there exists some $\varphi , \psi \in H^{\infty }$ and $f \in L^{\infty }$ such that

$$\begin{aligned} P_{{\mathcal {M}}}= \left[ \begin{array}{ccccc} t_{\varphi } &{}0\\ b_{{\bar{u}}^2f} &{} S_{\psi } \end{array}\right] . \end{aligned}$$

Since $P_{{\mathcal {M}}}$ is an orthogonal projection, we have $P_{{\mathcal {M}}}^{*}=P_{{\mathcal {M}}}$ and $P_{{\mathcal {M}}}^2=P_{{\mathcal {M}}}$ , it follows that

$$\begin{aligned} \left[ \begin{array}{ccccc} t_{\varphi } &{}0\\ b_{{\bar{u}}^2f} &{} S_{\psi } \end{array}\right] ^{*}&=\left[ \begin{array}{ccccc} t_{\varphi } &{}0\\ b_{{\bar{u}}^2f} &{} S_{\psi } \end{array}\right] , \end{aligned}$$

(22)

$$\begin{aligned} \left[ \begin{array}{ccccc} t_{\varphi } &{}0\\ b_{{\bar{u}}^2f} &{} S_{\psi } \end{array}\right] ^2&= \left[ \begin{array}{ccccc} t_{\varphi } &{}0\\ b_{{\bar{u}}^2f} &{} S_{\psi } \end{array}\right] . \end{aligned}$$

(23)

By Eq. (22), we get the following equations:

$$\begin{aligned} t_{\varphi }^{*}&=t_{\varphi },\end{aligned}$$

(24)

$$\begin{aligned} b_{{\bar{u}}^2f}^{*}&=0,\end{aligned}$$

(25)

$$\begin{aligned} b_{{\bar{u}}^2f}&=0, \end{aligned}$$

(26)

$$\begin{aligned} S_{\psi }^{*}&=S_{\psi }. \end{aligned}$$

(27)

Since $t_{\varphi }^{*}= t_{{\bar{\varphi }}}$ and $S_{\psi }^{*} =S_{{\bar{\psi }}}$, the equalities (24) and (27) imply that $\varphi$ and $\psi$ are real constants for $\varphi , \psi \in H^{\infty }$.

It follows that there appear four cases: both constants are non-zero, or one is zero, or both are zero, which give that the corresponding invariant subspaces of $D_z$ are the entire space, $uH^2$ or $\overline{zH^{2}}$, and $\{0\}$. Therefore, the result is desired.

(2) If $u(0) \ne 0$, by Theorem 1, there exists some $\varphi \in L^{\infty }$ such that

$$\begin{aligned} P_{{\mathcal {M}}}= \left[ \begin{array}{ccccc} t_{\varphi } &{} {\bar{u}}(0)b_{\overline{u\varphi }}^{*}\\ \frac{1}{{\bar{u}}(0)}b_{{\bar{u}}\varphi } &{} S_{\varphi } \end{array}\right] . \end{aligned}$$

Since $P_{{\mathcal {M}}}$ is an orthogonal projection, we have $P_{{\mathcal {M}}}^{*}=P_{{\mathcal {M}}}$ and $P_{{\mathcal {M}}}^2=P_{{\mathcal {M}}}$ , it follows that

$$\begin{aligned}&\left[ \begin{array}{ccccc} t_{\varphi } &{} {\bar{u}}(0)b_{\overline{u\varphi }}^{*}\\ \frac{1}{{\bar{u}}(0)}b_{{\bar{u}}\varphi } &{} S_{\varphi } \end{array}\right] ^{*}&=\left[ \begin{array}{ccccc} t_{\varphi } &{} {\bar{u}}(0)b_{\overline{u\varphi }}^{*}\\ \frac{1}{{\bar{u}}(0)}b_{{\bar{u}}\varphi } &{} S_{\varphi } \end{array}\right] , \end{aligned}$$

(28)

$$\begin{aligned}&\left[ \begin{array}{ccccc} t_{\varphi } &{} {\bar{u}}(0)b_{\overline{u\varphi }}^{*}\\ \frac{1}{{\bar{u}}(0)}b_{{\bar{u}}\varphi } &{} S_{\varphi } \end{array}\right] ^2&= \left[ \begin{array}{ccccc} t_{\varphi } &{} {\bar{u}}(0)b_{\overline{u\varphi }}^{*}\\ \frac{1}{{\bar{u}}(0)}b_{{\bar{u}}\varphi } &{} S_{\varphi } \end{array}\right] . \end{aligned}$$

(29)

By equation (28), we get

$$\begin{aligned} \left[ \begin{array}{ccccc} t_{\varphi }^{*} &{}\frac{1}{u(0)}b_{{\bar{u}}\varphi }^{*}\\ u(0)b_{\overline{u\varphi }}&{} S_{\varphi }^{*} \end{array}\right]&=\left[ \begin{array}{ccccc} t_{\varphi } &{} {\bar{u}}(0)b_{\overline{u\varphi }}^{*}\\ \frac{1}{{\bar{u}}(0)}b_{{\bar{u}}\varphi } &{} S_{\varphi } \end{array}\right] , \end{aligned}$$

(30)

which gives the following four equations:

$$\begin{aligned} t_{\varphi }^{*}&=t_{\varphi }, \end{aligned}$$

(31)

$$\begin{aligned} \frac{1}{u(0)}b_{{\bar{u}}\varphi }^{*}&={\bar{u}}(0)b_{\overline{u\varphi }}^{*}, \end{aligned}$$

(32)

$$\begin{aligned} u(0)b_{\overline{u\varphi }}&=\frac{1}{{\bar{u}}(0)}b_{{\bar{u}}\varphi }, \end{aligned}$$

(33)

$$\begin{aligned} S_{\varphi }^{*}&=S_{\varphi }. \end{aligned}$$

(34)

Since $t_{\varphi }^{*}= t_{{\bar{\varphi }}}$ and $S_{\varphi }^{*} =S_{{\bar{\varphi }}}$, both Eqs. (31) and (34) imply that $\varphi$ is real-valued. On the other hand, by Eqs.  (32), (33) and the fact that $\varphi$ is real-valued, we obtain

$$\begin{aligned} b_{{\bar{u}}\varphi }^{*}=|u(0)|^2b_{{\bar{u}}\varphi }^{*}, \, \, b_{{\bar{u}}\varphi }=|u(0)|^2b_{{\bar{u}}\varphi }. \end{aligned}$$

Since $u(0) \ne 0$, it follows that $b_{{\bar{u}}\varphi }=b_{{\bar{u}}\varphi }^{*}=0$. Furthermore, Eq. (29) and $b_{{\bar{u}}\varphi }=b_{{\bar{u}}\varphi }^{*}=0$ give that

$$\begin{aligned} \left[ \begin{array}{ccccc} t_{\varphi }^2 &{} 0\\ 0 &{} S_{\varphi }^2 \end{array}\right] = \left[ \begin{array}{ccccc} t_{\varphi } &{} 0\\ 0&{} S_{\varphi } \end{array}\right] ; \end{aligned}$$

thus we have $t_{\varphi }^2 = t_{\varphi }$ and $S_{\varphi }^2 = S_{\varphi };$ hence $t_{\varphi }=0$ or $I_{uH^2}$. Thus $P_{{\mathcal {M}}}=0$ or $I_{(K_u^2)^{\bot }}$, which means the possible reducing subspace for $D_z$ is either 0 or $(K_u^2)^{\bot }$. Therefore, $D_z$ has no nontrivial reducing subspaces if $u(0) \ne 0$. $\square$

Suppose u and v are nonconstant inner functions; define $D_z^u: (K_u^2)^{\bot } \rightarrow (K_u^2)^{\bot }$ by $D_z^ux_1=(uP{\bar{u}}+P_{-})zx_1$ for any $x_1 \in (K_u^2)^{\bot }$, and $D_z^v: (K_v^2)^{\bot } \rightarrow (K_v^2)^{\bot }$ by $D_z^vx_2=(vP{\bar{v}}+P_{-})zx_2$ for any $x_2 \in (K_v^2)^{\bot }$. Define ${\mathcal {J}}: (K_u^2)^{\bot } \rightarrow (K_v^2)^{\bot }$ by ${\mathcal {J}}=U_{v}U_{u}^{-1}$, where $U_{u}$ is the operator U introduced in the beginning of Sect. 2 and for which we add the indices to emphasize the inner functions in play. It is obvious that ${\mathcal {J}}$ is a unitary operator from $(K_u^2)^{\bot }$ to $(K_v^2)^{\bot }$. So $(K_u^2)^{\bot }$ and $(K_v^2)^{\bot }$ are isomorphic. Thus it is natural to ask whether the dual truncated Toeplitz operators $D_z^u$ and $D_z^v$ are unitarily equivalent. However, the following result shows that, in general, $D_z^u$ and $D_z^v$ are not unitarily equivalent:

Corollary 2

Let u and v be nonconstant inner functions with $u(0)=0$ and $v(0) \ne 0$, then $D_z^u$ and $D_z^v$ are not unitarily equivalent.

Proof

Assume $D_z^u$ and $D_z^v$ are unitarily equivalent; then there exists a unitary operator ${\mathcal {J}}_{u,v}: (K_u^2)^{\bot } \rightarrow (K_v^2)^{\bot }$ such that $D_z^u={\mathcal {J}}_{u,v}^{*}D_z^v{\mathcal {J}}_{u,v}.$ By Corollary 1, ${\mathcal {M}}=uH^2$ is a nontrivial reducing subspace for $D_z^u$, that is, $D_z^u{\mathcal {M}} \subset {\mathcal {M}}$ and $(D_z^u)^{*}{\mathcal {M}} \subset {\mathcal {M}}$, it follows that $({\mathcal {J}}_{u,v}^{*}D_z^v{\mathcal {J}}_{u,v}){\mathcal {M}} \subset {\mathcal {M}}$ and $({\mathcal {J}}_{u,v}^{*}D_z^v{\mathcal {J}}_{u,v})^{*}{\mathcal {M}} \subset {\mathcal {M}}$. Thus we have $D_z^v({\mathcal {J}}_{u,v}{\mathcal {M}}) \subset {\mathcal {J}}_{u,v}{\mathcal {M}}$ and $(D_z^v)^{*}({\mathcal {J}}_{u,v}{\mathcal {M}}) \subset {\mathcal {J}}_{u,v}{\mathcal {M}}$, that is, ${\mathcal {J}}_{u,v}{\mathcal {M}}$ is a nontrivial reducing subspace for $D_z^v$. However, $D_z^v$ has no nontrivial reducing subspace by Corollary 1, hence the assumption fails, thus this proves the result. $\square$